Question

help with implicit differentiation, please
(x^2y+xy)^2 (4xy) = 0

Answer 1

Please check working before use as simplification is complicated

Answer 2

thank you

Answer 3

Take the total derivative of(x^2y+xy)^2 (4xy) = 0:\[4 y \left(x y+x^2 y\right)^2 \text{Dt}[x]+4 x \left(x y+x^2 y\right)^2 \text{Dt}[y]+ \]\[8 x y \left(x y+x^2 y\right) \left(y \text{Dt}[x]+2 x y \text{Dt}[x]+x \text{Dt}[y]+x^2 \text{Dt}[y]\right)= 0\]Solve the above for Dt}[y]:\[\text{Dt}[y]=-\frac{y (3 \text{Dt}[x]+5 x \text{Dt}[x])}{3 x (1+x)} \]Factor out Dt[x] and divide both sides of the equation by Dt[x].\[y'=-\frac{(5 x+3) y}{3 x (x+1)} \]